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As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper.

For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$, you assumed $X$ was affine, but you don't need it for this). This condition is basically that the non-singular $X$ has rational singularities. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic http://front.math.ucdavis.edu/0911.3599.

For the second question "fibres are not unions of projective spaces", you've already seen one non-normal example. Ulrich gives a nice example of a non-rational component in the comment below (originally I said something stupid here).

The last question, "is it always true that $H^n(Y,O_Y)=0$?", here's an example which this time is normal. It is Section III of Cutkosky's A new characterization of rational surface singularities Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in.

Note: In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this answer on math overflow.

As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper.

For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$, you assumed $X$ was affine, but you don't need it for this). This condition is basically that the non-singular $X$ has rational singularities. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic https://arxiv.org/abs/0911.3599.

For the second question "fibres are not unions of projective spaces", you've already seen one non-normal example. Ulrich gives a nice example of a non-rational component in the comment below (originally I said something stupid here).

The last question, "is it always true that $H^n(Y,O_Y)=0$?", here's an example which this time is normal. It is Section III of Cutkosky's A new characterization of rational surface singularities Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in.

Note: In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this answer on math overflow.

As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper.

For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$, you assumed $X$ was affine, but you don't need it for this). This condition is basically that the non-singular $X$ has rational singularities. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic http://front.math.ucdavis.edu/0911.3599.

For the second question "fibres are not unions of projective spaces", you've already seen one non-normal example. Ulrich gives a nice example of a non-rational component in the comment below (originally I said something stupid here).

The last question, "is it always true that $H^n(Y,O_Y)=0$?", here's an example which this time is normal. It is Section III of Cutkosky's A new characterization of rational surface singularities Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in.

Note: In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this answer on math overflow.

As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper.

For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$, you assumed $X$ was affine, but you don't need it for this). This condition is basically that the non-singular $X$ has rational singularities. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic http://front.math.ucdavis.edu/0911.3599.

For the second question "fibres are not unions of projective spaces", you've already seen one non-normal example. Ulrich gives a nice example of a non-rational component in the comment below (originally I said something stupid here).

The last question, "is it always true that $H^n(Y,O_Y)=0$?", here's an example which this time is normal. It is Section III of Cutkosky's A new characterization of rational surface singularities Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in.

Note: In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this answer on math overflow.

As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper.

For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$, you assumed $X$ was affine, but you don't need it for this). This condition is basically that the non-singular $X$ has rational singularities. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic http://front.math.ucdavis.edu/0911.3599.

For the second question "fibres are not unions of projective spaces", you've already seen one non-normal example. I believe that fibers of such maps should always have unirational irreducible components (unirational means birational to an image of $\mathbb{A}^n$ under some map) but nothing more. I'm sure one can cook up normal examples too, but I don't know of any with the possible exception of the following counter-example to your third question (which I haven't checked).

The last question, "is it always true that $H^n(Y,O_Y)=0$?", here's an example which this time is normal. It is Section III of Cutkosky's A new characterization of rational surface singularities Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in.

Note: In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this answer on math overflow.

As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper.

For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$, you assumed $X$ was affine, but you don't need it for this). This condition is basically that the non-singular $X$ has rational singularities. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic http://front.math.ucdavis.edu/0911.3599.

For the second question "fibres are not unions of projective spaces", you've already seen one non-normal example. Ulrich gives a nice example of a non-rational component in the comment below (originally I said something stupid here).

The last question, "is it always true that $H^n(Y,O_Y)=0$?", here's an example which this time is normal. It is Section III of Cutkosky's A new characterization of rational surface singularities Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in.

Note: In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this answer on math overflow.